# Limit of Expectation values involving exponential i.i.d random variables

Let $$(X_n)_{n \in \mathbb{N}}$$ be a sequence of independent and identically distributed (i.i.d) random variables with $$\mathbb{E}[X_1] = 1$$ and $$\mathbb{V}[X_1] = 1$$. Show that $$\lim_{n \rightarrow \infty} \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2n} \bigl(\sum^{n}_{k=1} (X_k-1) \bigr)^2 \Bigr) \Bigr] = \frac{1}{\sqrt{2}}.$$ I am stuck on this problem not knowing how to approach it. Is there a hidden application of some theorem or is the proof elementary?
I would appreciate any tips on how to tackle this limit.

Edit: Based on Oscar's answer I wish to write down my attempt.
Let $$\mu = \mathbb{E}[X_1] = 1$$ and $$\sigma^2 = \mathbb{V}[X_1] = 1>0$$, then we may examine the exponent and observe that $$-\frac{1}{2n} \Bigl(\sum^{n}_{k=1} (X_k-1) \Bigr)^2 = -\frac{1}{2} \Bigl( \frac{\sum^{n}_{k=1}X_k-n}{\sqrt{n}} \Bigr)^2 = -\frac{1}{2} \Bigl( \frac{\sum^{n}_{k=1}X_k-n \mu}{\sqrt{n} \sigma} \Bigr)^2 =: -\frac{1}{2} Y_n^2$$ where we define the standardized variable $$Y_n = \frac{\sum^{n}_{k=1}X_k-n \mu}{\sqrt{n} \sigma}$$ for $$n \in \mathbb{N}$$.
The Central Limit Theorem is applicable and yields $$Y_n \rightarrow Y \sim \mathcal{N}(0,1)$$ in distribution, i.e. $$F_{Y_n}(t) \rightarrow F_Y(t) \, \, (n \rightarrow \infty)$$ for all $$t \in \mathbb{R}$$.
Now, we calculate the expectation value: $$\mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2} Y^2 \Bigr) \Bigr] = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp \Bigl(-\frac{1}{2} y^2 \Bigr) p(y) dy = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp \Bigl(-\frac{1}{2} y^2 \Bigr) \exp \Bigl(-\frac{1}{2} y^2 \Bigr) dy = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp \Bigl(-y^2 \Bigr) dy = \frac{1}{\sqrt{2\pi}} \sqrt{\pi} = \frac{1}{\sqrt{2}}$$

In order to finish the proof we need the statement: $$\lim_{n \rightarrow \infty} \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2} Y^2_n \Bigr) \Bigr] = \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2} Y^2 \Bigr) \Bigr].$$

Questions: Does this hold true?
And in general: If $$f: \mathbb{R} \rightarrow \mathbb{R}$$ is a Borel-measurable function, when does $$\lim_{n \rightarrow \infty} \mathbb{E} \Bigl[ f(Y_n) \Bigr] = \mathbb{E} \Bigl[ f(Y) \Bigr]$$ hold. What conditions must $$f$$ satisfy?

• Recall that you may rewrite the exponential of a sum as a product of exponentials. Moreover, the expectation of a product of independent r.v.'s is the product of the respective expectations. Commented Aug 3 at 10:22
• Actually, it may be more direct to just apply the CLT Commented Aug 3 at 10:43
• @tom If you know about Skorokhod's Representation theorem, then you'll know that if $X_{n}\xrightarrow{d}X$, then you can find representatives $Y_{n}$ and $Y$ such that $X_{n}=Y_{n}$ in distribution and $Y=X$ and $Y_{n}\xrightarrow{a.s.}Y$ almost surely. Now, that you know this, if you think about it, for $E(f(X_{n}))=E(f(Y_{n}))$ to converge to $E(f(X))=E(f(Y))$, it is reasonable to assume that $f(Y_{n})\xrightarrow{a.s.}f(Y)$ and $f(Y_{n})$ satisfies the DCT condition(or more generally, uniform integrability). One easy way to ensure both is to assume $f$ is continuous and bounded. Commented Aug 3 at 16:41
• @Tom Lucas Your attempt looks good to me, I'll add an edit to my answer to justify that final step. Please accept the answer if it's clear Commented Aug 22 at 16:03

Note that $$-\frac{1}{2n} \Big( \sum_{k=1}^n (X_k - 1) \Big)^2 = -\frac{1}{2} \Big( \sum_{k=1}^n \frac{(X_k - 1)}{\sqrt{n}} \Big)^2 = -\frac{1}{2} \Big( \sqrt{n} \sum_{k=1}^n \frac{(X_k - 1)}{n} \Big)^2,$$ and by the Central Limit theorem you have that $$\sqrt{n} \sum_{k=1}^n \frac{(X_k - 1)}{n} \overset{d}{\to} \mathcal{N}(0,1).$$ Also, recall moment generating functions.

--- Edit ---

To justify the final step you can use the following result:

Let $$X_1,X_2,\dots$$ be a sequence of random variables and suppose that $$X_n \to X$$ in distribution as $$n\to \infty$$. If $$h$$ is a real valued or complex valued, bounded, continuous function, then $$\mathbb{E}[h(X_n)] \to \mathbb{E}[h(X)] \quad \text{as} \ n \to \infty.$$

You can find a proof of this result in Allan Gut's Probability: A Graduate Course, Theorem 5.7. In your case, $$h = e^{-x^2/2}$$, which is of course bounded and continuous.

• Thank you. I edited my question to include a proof attempt and to clarify further problems. Commented Aug 3 at 15:18

Let $$S_{n}=\sum_{k=1}^{n}(X_{k}-1)$$.

Then $$S_{n}$$ is the sum of iid mean $$0$$ random variables with variance $$1$$.

The Central Limit Theorem implies $$\frac{S_{n}}{\sqrt{n}}\xrightarrow{d}N(0,1)$$.

One very useful, and often the standard characterization of convergence in distribution is the following functional analytic one:-

We say $$Z_{n}\xrightarrow{d}Z$$ if for all $$f\in C_{b}(\Bbb{R})$$, i.e. the space of bounded continuous functions of $$\Bbb{R}$$, we have $$E(f(Z_{n}))\xrightarrow{n\to\infty}E(f(Z))$$

Now, define $$\displaystyle f(x)=e^{-\frac{x^{2}}{2}}$$ for $$x\in\Bbb{R}$$ . It is obvious to see that $$f$$ is a continuous and bounded function.

Now, we set $$Z_{n}=\frac{S_{n}}{\sqrt{n}}$$ and $$Z=N(0,1)$$.

So, the above characterization says precisely that:-

$$\mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2n} \bigl(\sum^{n}_{k=1} (X_k-1) \bigr)^2 \Bigr) \Bigr]=\mathbb{E}\bigg(\frac{-S_{n}^{2}}{2n}\bigg)=E(f(Z_{n}))\to E(f(Z))$$

So all you need to do is compute that

$$E(f(Z))=\int_{-\infty}^{\infty}e^{-\frac{z^{2}}{2}}\cdot\dfrac{e^{-\frac{z^{2}}{2}}}{\sqrt{2\pi}}\,dz =\frac{1}{\sqrt{2}}$$